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gulaghasi [49]
3 years ago
11

Given that 16 ounces equals 1 pound, how many ounces are in 8.9 pounds

Mathematics
2 answers:
kondaur [170]3 years ago
8 0
The answer is B - 142.4

pishuonlain [190]3 years ago
5 0
Hi,
About 142.4 is the answer!

I hope this helped!
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Thinking The dot plot shows
dedylja [7]

Answer:

01001000 01100001 01101000 01100001 01101000 01100001 00100000 01111001 01101111 01110101 00100000 01101010 01110101 01110011 01110100 00100000 01100111 01101111 01110100 00100000 01110011 01100011 01100001 01101101 01101101 01100101 01100100 00100000 01100010 01111001 00100000 01101101 01100101 00100000 01001001 01101101 00100000 01110100 01101111 00100000 01100101 01110000 01101001 01101011 00100000 00111011 00101001 00100000 01101000 01100001 01101000 01100001 01101000 01100001 01101000 01100001 01101000 01100001 01101000 01100001 01101000 01100001 01101000 01100001 01101000 01100001 01101000 01100001 01101000 01100001 01101000 01100001 01101000 00100000 01001001 00100000 01001100 01001001 01001011 01000101 00100000 01011001 01000001 01000011 01010101 01010100 00100000 01000111 00100000 00101010 01010011 01001100 01000001 01010000 00100000 01000001 01001000 01001000 01001000 01001000 01001000 01001000 01001000 01001000 01001000 01001000 01001000 01001000 01001000 01001000 01001000 01001000 01001000 01001000 01001000 01001000 01001000

Step-by-step explanation:

8 0
3 years ago
(a-b)^2/(1/a-1/b) simplify
leva [86]

Final result :

(b - a) • (a2 + ab + b2)

————————————————————————

a2b3

Step by step solution :

Step 1 :

1

Simplify —

a

Equation at the end of step 1 :

1 1 1

————-———— ÷ (—•b)

(a2) (b2) a

Step 2 :

1

Simplify ——

b2

Equation at the end of step 2 :

1 1 b

———— - —— ÷ —

(a2) b2 a

Step 3 :

1 b

Divide —— by —

b2 a

3.1 Dividing fractions

To divide fractions, write the divison as multiplication by the reciprocal of the divisor :

1 b 1 a

—— ÷ — = —— • —

b2 a b2 b

Multiplying exponential expressions :

3.2 b2 multiplied by b1 = b(2 + 1) = b3

Equation at the end of step 3 :

1 a

———— - ——

(a2) b3

Step 4 :

1

Simplify ——

a2

Equation at the end of step 4 :

1 a

—— - ——

a2 b3

Step 5 :

Calculating the Least Common Multiple :

5.1 Find the Least Common Multiple

The left denominator is : a2

The right denominator is : b3

Number of times each Algebraic Factor

appears in the factorization of:

Algebraic

Factor Left

Denominator Right

Denominator L.C.M = Max

{Left,Right}

a 2 0 2

b 0 3 3

Least Common Multiple:

a2b3

Calculating Multipliers :

5.2 Calculate multipliers for the two fractions

Denote the Least Common Multiple by L.C.M

Denote the Left Multiplier by Left_M

Denote the Right Multiplier by Right_M

Denote the Left Deniminator by L_Deno

Denote the Right Multiplier by R_Deno

Left_M = L.C.M / L_Deno = b3

Right_M = L.C.M / R_Deno = a2

Making Equivalent Fractions :

5.3 Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example : 1/2 and 2/4 are equivalent, y/(y+1)2 and (y2+y)/(y+1)3 are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

L. Mult. • L. Num. b3

—————————————————— = ————

L.C.M a2b3

R. Mult. • R. Num. a • a2

—————————————————— = ——————

L.C.M a2b3

Adding fractions that have a common denominator :

5.4 Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

b3 - (a • a2) b3 - a3

————————————— = ———————

a2b3 a2b3

Trying to factor as a Difference of Cubes:

5.5 Factoring: b3 - a3

Theory : A difference of two perfect cubes, a3 - b3 can be factored into

(a-b) • (a2 +ab +b2)

Proof : (a-b)•(a2+ab+b2) =

a3+a2b+ab2-ba2-b2a-b3 =

a3+(a2b-ba2)+(ab2-b2a)-b3 =

a3+0+0+b3 =

a3+b3

Check : b3 is the cube of b1

Check : a3 is the cube of a1

Factorization is :

(b - a) • (b2 + ab + a2)

Trying to factor a multi variable polynomial :

5.6 Factoring b2 + ab + a2

Try to factor this multi-variable trinomial using trial and error

Factorization fails

Final result :

(b - a) • (a2 + ab + b2)

————————————————————————

a2b3

4 0
4 years ago
Sam is rowing a boat away from a dock. The graph shows the relationship
yan [13]

Answer:

Step-by-step explanation:

B

5 0
3 years ago
Read 2 more answers
A point having a negative abscissa and negative ordinate is in quadrant ____.
Aloiza [94]
Your answer would be Quadrant III (Answer #3)

5 0
3 years ago
Read 2 more answers
M&M plain candies come in various colors. According to the M&M/Mars Department of Consumer Affairs, the distribution of
lyudmila [28]

Answer:

A) 0.12. Yes. Choosing a green and blue M&M is possible

B) 0.43. Yes. Choosing a yellow and red M&M is possible

C) 0.78

Step-by-step explanation:

First of all, the summation of the distribution of all colours is;

Σ(all colors ) = 22% + 20% + 23% + 10% + 6% + 6% + 13% = 100%, or 1.

Thus;

a) P(green candy or blue candy) is;

P(GREEN ∪ BLUE) = P(G) + P(BL)

P(GREEN ∪ BLUE) = 6%+6%

P(GREEN ∪ BLUE) = 12% or 0.12

Now, due to the fact that we have to choose ONE candy and only ONE candy at random, then they are mutually exclusive: Yes. Choosing a green and blue M&M is possible

b)P(yellow candy or red candy is;

P(YELLOW ∪ RED) = P(Y) + P(R)

P(YELLOW ∪ RED) = 20% + 23% = 43% or 0.43

Yes. Choosing a yellow and red M&M is possible

c) P(NOT PURPLE)

the probability of having a purple is;

P(PURPLE) = 22% or 0.22

So, the Probability of NOT having a PURPLE is 1 - 0.22 = 0.78

4 0
3 years ago
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